Sandra is 36 years old. Her son is 14 years old. How many years ago was Sandra 3 times older than her son? years ago

Answers

Answer 1

Answer: 36
how many years ago (x years) was sandra (36-x) 3 times as old as son (at that time, 14-x)

sandra's age x years ago is 3 times son's age x years ago
36-x=3(14-x)
solve
distribute
36-x=42-3x
add 3x both sides
36+2x=42
minus 36
2x=6
divide 2
x=3

3 years ago
(36-3=33, 14-3=11, 11*3=33)

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A furniture maker uses the specification 21.88< or = w

Answers

Seems like your question is quite incomplete, I bet here is the missing part *<or = 22.12 for the width w in inches of a desk drawer.
According to the fact that abs(w-22) should be less than equal to 12, the answer should have a look like this : Iw - 22I < or = 0.12. I am pretty sure it will help you.
Next time check it carefully. Regards!

If x is a real number such that x³ = 64, then x² + √x =?

Answers

$x^3 = 64\n \nx=\sqrt[3]{64}\n \nx=4 \n \n \n x^2 + √( x )=4^2+√(4)=16+2 =18$

If x³=64
,x=∛64=4
x²=16
√x=4
x²+√x=16+2
⇒18

Which sentence correctly describes a data set that follows a normal distribution with a standard deviation of 4 and a mean of 14?68% of the data points lie between 10 and 14.

68% of the data points lie between 8 and 12.

68% of the data points lie between 10 and 18.

68% of the data points lie between 10 and 16.

Answers

Answer:

Option (c) is correct.

68% of the data points lie between 10 and 18.

Step-by-step explanation:

Given : a normal distribution with a standard deviation of 4 and a mean of 14

We have to choose the sentence that correctly describes a data set that follows a normal distribution with a standard deviation of 4 and a mean of 14.

Since, given 68% data.

We know mean of data lies in middle.

And standard deviation is distribute equally about the mean that is 50% of values less than the mean and 50% greater than the mean.

So, 68% of data lies

mean - standard deviation = 14 - 4 = 10

mean + standard deviation = 14 + 4 = 18

So, 68% of the data points lie between 10 and 18.

"For any normal distribution, approximately 68% of the distribution will lie within one standard deviation of the mean.
That means that, for this distribution, 68% of it will lie between 14-4=10 and 14+4=18."

If y varies directly as x, and y is 6 when x is 72, which is the value of y when x is 8 ?

Answers

$(y)/(x) = k$

$(6)/(72) = k$

$(1)/(12) = k$

********************************

$(y)/(x) = k$

$(y)/(8) = (1)/(12)$

12y = 8

$y = (8)/(12)$

$y = (2)/(3)$

y = $(2)/(3)$

given y ∝ x

To convert to an equation multiply by k the constant of variation

y = kx

to find k use the given condition y = 6 when x = 72

y = kx ⇒ k = $(y)/(x)$ = $(6)/(72)$ = $(1)/(12)$

equation is → y = $(1)/(12)$ x

when x = 8 then

y = $(1)/(12)$ × 8 = $(8)/(12)$ = $(2)/(3)$

Verify the basic identity. What is the domain of validity? cot theta = cos theta csc theta

Answers

Both sides can be the domain of validity since both are just simple but what we are going to change is the right side.
Let us review that $cot \alpha = (cos \alpha )/(sin \alpha )$ and $csc \alpha = (1)/(sin \alpha )$ .
So, to prove the following identity:
$cot \alpha =cos \alpha csc \alpha$
Let us substitute the value of csc with respect to sin.
$cot \alpha =cos \alpha * (1)/(sin \alpha )$
$cot \alpha = (cos \alpha )/(sin \alpha )$
$cot \alpha =cot \alpha$